To find out how many grams of hydrogen bromide (HBr) is formed when 2.23 grams of hydrogen (H₂) reacts completely with bromine (Br₂), follow these steps:
1. Calculate the moles of hydrogen gas (H₂):
- The molar mass of hydrogen (H₂) is 2.02 g/mol.
- Use the formula for moles:
[tex]\[
\text{moles of H}_2 = \frac{\text{mass of H}_2}{\text{molar mass of H}_2}
\][/tex]
Substituting the given values:
[tex]\[
\text{moles of H}_2 = \frac{2.23 \, \text{g}}{2.02 \, \text{g/mol}} \approx 1.104 \, \text{mol}
\][/tex]
2. Stoichiometry of the reaction:
- The balanced chemical reaction is:
[tex]\[
H_2 + Br_2 \rightarrow 2HBr
\][/tex]
- According to the balanced equation, 1 mole of hydrogen gas (H₂) produces 2 moles of hydrogen bromide (HBr).
3. Calculate the moles of hydrogen bromide (HBr) produced:
- Since 1 mole of H₂ produces 2 moles of HBr,
[tex]\[
\text{moles of HBr} = 2 \times \text{moles of H}_2 = 2 \times 1.104 \approx 2.208 \, \text{mol}
\][/tex]
4. Calculate the mass of hydrogen bromide (HBr):
- The molar mass of hydrogen bromide (HBr) is calculated by adding the atomic masses of hydrogen and bromine:
[tex]\[
\text{molar mass of HBr} = 1.008 \, \text{g/mol} + 79.9 \, \text{g/mol} \approx 80.9 \, \text{g/mol}
\][/tex]
- Use the formula for mass:
[tex]\[
\text{mass of HBr} = \text{moles of HBr} \times \text{molar mass of HBr}
\][/tex]
Substituting the calculated moles of HBr:
[tex]\[
\text{mass of HBr} = 2.208 \, \text{mol} \times 80.9 \, \text{g/mol} \approx 178.62 \, \text{g}
\][/tex]
Therefore, the closest answer, considering significant figures, is approximately 179 grams of HBr (rounded to the nearest whole number). Hence, the best answer is:
B. 179 g of HBr
